Which of the following numbers is a multiple of 2? ${73,81,99,112,115}$
Answer: The multiples of $2$ are $2$ $4$ $6$ $8$ ..... In general, any number that leaves no remainder when divided by $2$ is considered a multiple of $2$ We can start by dividing each of our answer choices by $2$ $73 \div 2 = 36\text{ R }1$ $81 \div 2 = 40\text{ R }1$ $99 \div 2 = 49\text{ R }1$ $112 \div 2 = 56$ $115 \div 2 = 57\text{ R }1$ The only answer choice that leaves no remainder after the division is $112$ $ 56$ $2$ $112$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $2$ are contained within the prime factors of $112$ $112 = 2\times2\times2\times2\times7 2 = 2$ Therefore the only multiple of $2$ out of our choices is $112$. We can say that $112$ is divisible by $2$.